5.4 Indefinite Integrals and the Net Change Theorem/3: Difference between revisions

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& \frac{d}{dx} {[\sin{x} - \frac{1}{3} \sin^3{x} +c]} \\[2ex]
& \frac{d}{dx} {[\sin{x} - \frac{1}{3} \sin^3{x} +c]} \\[2ex]


& {\cos{x} - \frac{1}{3}\cdot 3\sin{x^2} \cos{x} +0} \\[2ex]
& ={\cos{x} - \frac{1}{3}\cdot 3\sin{x^2} \cos{x} +0} \\[2ex]


& \cos{x} - \sin^2{x}\cos{x} \\[2ex]
& =\cos{x} - \sin^2{x}\cos{x} \\[2ex]


& = \cos^3{x}




\end{align}
\end{align}
</math>
</math>

Revision as of 19:30, 26 August 2022

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}&\int \cos ^{3}xdx=\sin {x}-{\frac {1}{3}}\sin ^{3}x+C\\[2ex]&{\frac {d}{dx}}{[\sin {x}-{\frac {1}{3}}\sin ^{3}{x}+c]}\\[2ex]&={\cos {x}-{\frac {1}{3}}\cdot 3\sin {x^{2}}\cos {x}+0}\\[2ex]&=\cos {x}-\sin ^{2}{x}\cos {x}\\[2ex]&=\cos ^{3}{x}\end{aligned}}}